Mysql 运行多线程代码规范时锁定等待超时

我有一个类，它在许多线程中将所有用户名更改为John

class Users::UpdateName
  def call
    User
      .unscoped
      .find_in_batches do |batch|
        update_batch(batch)
      end
  end

  def update_batch(users_batch)
    users_batch
      .each_slice(100)
      .map { |users| Thread.new { update_all(users) } }
      .each(&:join)
  end

  def update_all(users)
    users.each { |u| u.update_columns(first_name: 'John') }
  end
end

其中类用户是具有默认范围的活动记录模型

default_scope{where（归档：false）}

我用以下方法测试我的班级：

describe Users::UpdateName

  it 'should update first name of an archived user' do
    user = User.create(first_name: 'Jane', archived: true)

    Users::UpdateName.new.call

    expect(user.reload.first_name).to eq 'John'
  end

end

但是当我运行测试时，它失败了

 ActiveRecord::StatementInvalid:
   Mysql2::Error: Lock wait timeout exceeded; try restarting transaction: UPDATE `users` SET `users`.`first_name` = 'John' WHERE `users`.`id` = 10

当我用

查看数据库状态时，显示引擎innodb状态它给了我

TRANSACTIONS
------------
Trx id counter 4359880
Purge done for trx's n:o < 4359879 undo n:o < 0 state: running but idle
History list length 1034
LIST OF TRANSACTIONS FOR EACH SESSION:
---TRANSACTION 0, not started
MySQL thread id 198, OS thread handle 0x700005048000, query id 43397 localhost root init
show engine innodb status
---TRANSACTION 4359879, ACTIVE 13 sec starting index read
mysql tables in use 1, locked 1
LOCK WAIT 2 lock struct(s), heap size 360, 1 row lock(s)
MySQL thread id 200, OS thread handle 0x700004f7c000, query id 43396 localhost root updating
UPDATE `users` SET `users`.`first_name` = 'John' WHERE `users`.`id` = 12
------- TRX HAS BEEN WAITING 13 SEC FOR THIS LOCK TO BE GRANTED:
RECORD LOCKS space id 173721 page no 3 n bits 72 index `PRIMARY` of table `mysql_lock_wait_timeout_problem_test`.`users` trx id 4359879 lock_mode X locks rec but not gap waiting
Record lock, heap no 2 PHYSICAL RECORD: n_fields 7; compact format; info bits 0
 0: len 8; hex 800000000000000c; asc         ;;
 1: len 6; hex 0000004286c2; asc    B  ;;
 2: len 7; hex d6000001510110; asc     Q  ;;
 3: len 4; hex 4a616e65; asc Jane;;
 4: len 1; hex 81; asc  ;;
 5: len 5; hex 99a034ac56; asc   4 V;;
 6: len 5; hex 99a034ac56; asc   4 V;;

------------------
---TRANSACTION 4359874, ACTIVE 13 sec
2 lock struct(s), heap size 360, 1 row lock(s), undo log entries 1
MySQL thread id 199, OS thread handle 0x700004ef4000, query id 43393 localhost root
Trx read view will not see trx with id >= 4359879, sees < 4359879
--------
FILE I/O
--------
I/O thread 0 state: waiting for i/o request (insert buffer thread)
I/O thread 1 state: waiting for i/o request (log thread)
I/O thread 2 state: waiting for i/o request (read thread)
I/O thread 3 state: waiting for i/o request (read thread)
I/O thread 4 state: waiting for i/o request (read thread)
I/O thread 5 state: waiting for i/o request (read thread)
I/O thread 6 state: waiting for i/o request (write thread)
I/O thread 7 state: waiting for i/o request (write thread)
I/O thread 8 state: waiting for i/o request (write thread)
I/O thread 9 state: waiting for i/o request (write thread)
Pending normal aio reads: 0 [0, 0, 0, 0] , aio writes: 0 [0, 0, 0, 0] ,
 ibuf aio reads: 0, log i/o's: 0, sync i/o's: 0
Pending flushes (fsync) log: 0; buffer pool: 0
1320 OS file reads, 89466 OS file writes, 79251 OS fsyncs
0.00 reads/s, 0 avg bytes/read, 0.38 writes/s, 0.24 fsyncs/s
-------------------------------------

您可以了解如何初始化应用程序以在中重现错误。
您还可以找到所有的源代码
我可以用更简单的代码重现这个bug：

ActiveRecord::Base.transaction do
  user = User.create! first_name: 'Jane'

  t = Thread.new do
    user.update_columns first_name: 'John'
  end

  t.join
end

说明：

主线程在事务中创建一条记录，并将该记录标记为不可更新。同时，另一个线程尝试更新该记录并等待主事务。主事务无法完成，因为它正在等待另一个线程

其中类用户是一个活动记录模型，具有默认范围default_scope{where（归档：false）}

用户类中的默认作用域在那里不起任何作用
我无法重现这一点。我正在使用mysql2-0.5.1。你能试试使用这个版本的mysql gem吗？@RahulSharma好的，我会试试。我使用mysql2-v0.3.20。我更新了文件的要点。您是否尝试克隆github repo？@RahulSharma测试失败，在我的电脑上出现与mysql2-0.5.1相同的错误。是的，我克隆了github repo。能否检查本地计算机中的测试数据库是否已存在一些条目？
ActiveRecord::Base.transaction do
  user = User.create! first_name: 'Jane'

  t = Thread.new do
    user.update_columns first_name: 'John'
  end

  t.join
end